- #1

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and can anyone help with this one:

the integral from 0 to pi of (sinti+costj+tk)?

is it -costi + sintj +(t^2/2)k?

thank you!

- Thread starter don23
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- #1

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and can anyone help with this one:

the integral from 0 to pi of (sinti+costj+tk)?

is it -costi + sintj +(t^2/2)k?

thank you!

- #2

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Well e^-t x e^t is same as e^t-t so It should be e^0=1.don23 said:

and can anyone help with this one:

the integral from 0 to pi of (sinti+costj+tk)?

is it -costi + sintj +(t^2/2)k?

thank you!

I havent started on Integrals yet.

- #3

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I believe

[tex] e^t e^{ - t} = e^0 = 1 [/tex]

and

[tex]\int\limits_0^\pi {\left( {i\sin t + j\cos t + kt} \right)dt} = 2i + k\frac{{\pi ^2 }}{2}[/tex]

Just remember to take the limits of the integral.

You have antiderivative, now apply the algebra

[tex] e^t e^{ - t} = e^0 = 1 [/tex]

and

[tex]\int\limits_0^\pi {\left( {i\sin t + j\cos t + kt} \right)dt} = 2i + k\frac{{\pi ^2 }}{2}[/tex]

Just remember to take the limits of the integral.

You have antiderivative, now apply the algebra

Last edited:

- #4

VietDao29

Homework Helper

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bomba923, I think you've made a mistake here,[tex]\int\limits_0^\pi {\left( {i\sin t + j\cos t + kt} \right)dt} = 2i + k\frac{{\pi ^2 }}{2} + C [/tex]

First, I think the question reads:

[tex]\int_0 ^ \pi (\sin (it) + \cos (jt) + kt) dt[/tex]

And second, where does the little 'C' come from?

-------------------

Nope, it isn't.don23 said:is it -costi + sintj +(t^2/2)k?

In your post, you didn't state the variable of integration. So I assume it is dt.

So:

[tex]\int (\sin (it) + \cos (jt) + kt) dt = \int \sin (it) dt + \int \cos (jt) dt + \int kt dt[/tex]

[tex]= \int \frac{1}{i} \sin (it) d(it) + \int \frac{1}{j} \cos (jt) d(jt) + k \int t dt[/tex]

[tex]= \frac{1}{i} \int \sin (it) d(it) + \frac{1}{j} \int \cos (jt) d(jt) + k \int t dt = ...[/tex]

Can you go from here?

Viet Dao,

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- #5

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You have confused me here.VietDao29 said:First, I think the question reads:

[tex]\int_0 ^ \pi (\sin (it) + \cos (jt) + kt) dt[/tex]

I thought the question asked:

[tex]\int_0 ^ \pi ((\sin t)i + (\cos t)j + (t)k) dt[/tex]

My answer is

[tex] i + j + \frac{{\pi ^2 }}{2}k[/tex]

- #6

VietDao29

Homework Helper

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If I am asking some question like that, I am going to put theBuzzer said:You have confused me here.

I thought the question asked:

[tex]\int_0 ^ \pi ((\sin t)i + (\cos t)j + (t)k) dt[/tex]

Something like this:

i sint + j cost + kt.

And even if the question reads like this:Buzzer said:My answer is

[tex] i + j + \frac{{\pi ^2 }}{2}k[/tex]

[tex]\int_0 ^ \pi ((\sin t)i + (\cos t)j + (t)k) dt[/tex]

That is still

--------------

EDIT:

If the question happens to be:

int (i sint + j cost + kt) dt

You can use bomba923's suggestion. You should omit the constant of integration 'C' in his answer.

Viet Dao,

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- #7

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Ok, but I learnt to put i, j and k at the end when I did vectors....VietDao29 said:If I am asking some question like that, I am going to put thei, andjin front ofthe cos and sin functions.

Something like this:

i sint + j cost + kt.

I did it again and got the same answer. Where does 2i come from?

EDIT

How stupid of me, I had pi/2 in my head instead of pi.

Last edited:

- #8

HallsofIvy

Science Advisor

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Yes, the original post's "integral from 0 to pi of (sinti+costj+tk)"

[tex]\int_0^{\pi}{(sin t)i+ (cos t)j+ (t)k} dt[/itex]

to distinguish it form [tex]\int_0^{\pi}{sin ti+ cos tj+ tk} dt[/itex] but I think the "i, j, k" suggested vectors rather than numbers.

don23, It is true that

[tex]\int {(sin t)i+ (cos t)j+ (t)k}dt= -cos(t)i+ sin(t)j+ (\frac{1}{2}t^2+ C[/tex]

but you forgot to evaluate it between 0 and [itex]\pi[/itex].

[tex](-cos \pi)i+ (sin /pi)j+ \frac{1}{2}(/pi)^2= i+ 0j+ \frac{\pi^2}{2}k[/tex]

and

[tex](-cos 0)i+ (sin 0)i+ \frac{1}{2}(0)^2= -i+ 0j+ 0k[/tex]

so

[tex]\int_0^{\pi}{(sin t)i+ (cos t)j+ (t)k}dt= 2i +\frac{\pi^2}{2}k[/tex]

- #9

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Thank you Halls of Ivy

(you've vindicated me!)

(you've vindicated me!)

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